Explanation

(c) From
$R_{t} = R_{0}(1 + \alpha t)$
$5 = R_{o}(1 + 50\alpha)\ \ \ \ \ \ \ldots(i)$
$and\ \ \ \ 6 = R_{o}(1 + 100\alpha)\ \ldots(ii)$
$\therefore\frac{5}{6} = \frac{1 + 50\alpha}{1 + 100\alpha}$
$\Longrightarrow \alpha = \frac{1}{200}$
Putting value of $\alpha$in Eq. (i), we get
$\ \ \ \ \ \ 5 = R_{0}\left( 1 + 50 \times \frac{1}{200} \right)\ $
$\therefore R_{0} = 4\Omega$